Question

select *, 
   (select max(h.date) 
    from hour_cards as h  
    where h.subordinate_id = hour_cards.subordinate_id
      and h.date > hour_cards.date and 
   h.date - hour_cards.date <= INTERVAL 1 minute) as wrong_entry
   from hour_cards

我试图进行查询，如果他/她在委托或退出工作时读了两次或更多的卡，我会给每个工人的所有条目，但上面的查询给我一次错误， #1064 - You have an error in your SQL syntax; check the manual that corresponds to your MySQL server version for the right syntax to use near ') as wrong_entry from hour_cards

Answer 1

您只能在date calculation function中使用间隔。

您的查询必须如下：

select *, 
   (select max(h.date) 
    from hour_cards as h  
    where h.subordinate_id = hour_cards.subordinate_id
      and h.date > hour_cards.date and 
   DATE_ADD(h.date, INTERVAL 1 minute) <= hour_cards.date  as wrong_entry
   from hour_cards

Answer 2

您可以使用TIMESTAMPDIFF使其正确：

 TIMESTAMPDIFF(MINUTE,h.date,hour_cards.date) <= 1

mysql，查询间隔错误

2 个答案: